Consider \({\mathbb {R}}^{[0, 1]}\)  – the space of all functions \(f: [0, 1] \rightarrow {\mathbb {R}}\)  equipped with the standard product topology, i.e. the topology of pointwise convergence. The space \(X\)  we are looking for will be the subspace of \({\mathbb {R}}^{[0, 1]}\)  consisting of functions with countable support. In other words, \(f: [0, 1] \rightarrow {\mathbb {R}}\)  lies in \(X\)  if the set \({\text{supp}}f := \lbrace t \in [0, 1]: f(t) \ne 0\rbrace \)  is at most countable. Since \(X\)  is a dense proper subspace of \({\mathbb {R}}^{[0, 1]}\) , it cannot be complete (a complete subspace of any TVS is closed [1]}). Let us demonstrate that \(X\)  is countably complete.
Let \(\mathfrak {F}\)  be a filter on \({\mathbb {N}}\) , and \(x = (x_n) \in 2^X\)  be an \(\mathfrak {F}\) - Cauchy sequence in \(X\) . Then \(x\)  is \(\mathfrak {F}\) - Cauchy as a sequence in \({\mathbb {R}}^{[0, 1]}\) . By the completeness of \({\mathbb {R}}^{[0, 1]}\) , there is \(f \in {\mathbb {R}}^{[0, 1]}\)  such that \(f = \lim _\mathfrak {F}x_n\)  in \({\mathbb {R}}^{[0, 1]}\) .
